Find $\lim_{h\to 0}\dfrac{5\log(2+h)-5\log(2)}{h}$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac52$ (Choice B) B $\dfrac{5}{2\ln(10)}$ (Choice C) C $5\log(2)$ (Choice D) D The limit doesn't exist
Explanation: The limit expression has the following form: $\lim_{h\to 0}\dfrac{f(t+h)-f(t)}{h}$ Therefore, it describes a derivative of a certain function $f$ at a certain $x$ -value $t$. Can you recognize what $f$ and $t$ are? Since the numerator is $5\log(2+h)-5\log(2)$, we can tell that the function is $f(x)=5\log(x)$ and the $x$ -value is $2$. In other words, the limit expression is equal to $f'(2)$ for $f(x)=5\log(x)$. Let's find $f'(x)$ : $f'(x)=5\cdot \dfrac{1}{\ln(10)\cdot x}=\dfrac{5}{x\ln(10)}$ Now let's evaluate $f'(2)$ : $f'(2)=\dfrac{5}{2\ln(10)}$ In conclusion, $\lim_{h\to 0}\dfrac{5\log(2+h)-5\log(2)}{h}=\dfrac{5}{2\ln(10)}$.